Edit Distance

Edit Distance

Given two words word1 and word2, find the minimum number of steps required to convert word1 to word2. (each operation is counted as 1 step.)

You have the following 3 operations permitted on a word:

a) Insert a character
b) Delete a character
c) Replace a character

Solution 

This is a dynamic programming problem. Use a two-dimensional array dp[i][j] represent the minimum step needed to convert the 0 to i characters in string1 into 0 to j characters in string2. For example, if word1= "abc", word2 = "ab", then dp[1][2] represents the minimum step needed from converting a to ab. There are two conditions we need to take into account.

1. word1[i - 1] == word2[j - 1]

For example, word1 = "acc" and word2 = "abc".    dp[3][3] = dp[2][2].

2. word[i - 1] != word[j - 1] word1 = "ab" word2 = "ad". there are three sub conditions here:

      —insert

        ab —> a  then insert d        dp[2][2] = dp[2][1] + 1;

    —delete

        a —>ad then delete b        dp[2][2] = dp[1][2] + 1;

    —replace

        a —>a then replace b with d dp[2][2] = dp[1][1] + 1;

     Choose the minimum value of the three sub conditions.

public class Solution {    public int minDistance(String word1, String word2) {        int len1 = word1.length();        int len2 = word2.length();                int[][] dp = new int[len1 + 1][len2 + 1];        int min = Integer.MAX_VALUE;                for (int i = 0; i <= len1; i++) {            dp[i][0] = i;        }                for (int j = 0; j <= len2; j++) {            dp[0][j] = j;         }                for (int i = 1; i <= len1; i++) {            for (int j = 1; j <= len2; j++) {                if (word1.charAt(i - 1) == word2.charAt(j - 1)) {                    dp[i][j] = dp[i - 1][j - 1];                } else {                    min = Math.min(dp[i][j - 1] + 1, dp[i - 1][j] + 1);                    min = Math.min(min, dp[i - 1][j - 1] + 1);                    dp[i][j] = min;                }            }        }                        return dp[len1][len2];    }}